Investigation the Resonance and Oscillation - Lab Report Example

Lab report on the Resonance Experiment Your name Name of Assignment 9th October , 2010 ABSTRACT There has been a high rate of development and usage of oscillation in the recent past which has realized the design and modeling of machines with high-tech inbuilt capacities relating to their speed and frequencies without much jostling. This technological advancement has had a dark side since these refined high speeds have in turn led to increased cases of failures. Their results indicated some inception of rotating stall and decrease in flutter stability, though their findings is conclusive because available computational resources did not allow experiments data to be included. The aim of this paper is to conclude the work of experiment of the class, to investigate the resonance and oscillation. INTRODUCTION There are many researches that are currently underway regarding bird strikes since it is a phenomenon that has gripped airline operations management with a zest that requires their immediate attention and response to try and understand it and innovatively institute ways of improving it. In discussing these many areas of interest that experiments have been conducted, a detailed study of resonance and oscillation is necessary especially for the components that are adversely affected by it. EXPERIMENTAL DETAILS Explicit analysis was used and it is a numerical technique that is commonly used for the analysis of highly non-linear behaviour of materials with inelastic strains, high strain rates, and large deformations. The materials used in the experiment fitted into model the full non-linear behaviour to failure of the composite materials. This model is preferred since rather than using the ply material properties, the bi-phase model requires the constituent material properties and the fibre volume fraction to determine the stiffness of each ply. During high-speed impact, large strain distortions will inevitably occur, leading to a decrease in solution time-step and possible negative elemental volumes. The excessively distorted elements that possess negative volumes are then eliminated after each time step. However, this automatic elimination procedure usually introduces artificial oscillations in the contact force between the oscillation equipments. Fortunately, the strategy of employing highly refined meshes that encounter element elimination can be used to alleviate this problem. The bob’s mechanical property actually changes from the low-velocity to the high-velocity if pushed by force. It must be stressed that the computational requirement for such a calculation is very considerable since a time-accurate, nonlinear viscous analysis must be undertaken for a whole assembly because of the loss of symmetry in the flow. Experimental and predicted results were quoted over a wide range of impact velocities and energy levels for single bobs and rotor assemblies. It was shown that small fan blades experienced gross tensional untwist, followed by gross bending under medium and high-energy impacts. RESULTS When a pendulum swings back and forth, a string or thin rod constrains the bob to move along a circular are: However, for oscillations with small amplitude, we assume that the bob move back and forth along the x-axis; the vertical motion of the bob is negligible. This expressed as shown below; Since the weight of the bob has no x-component, the restoring force is the x-component of the force due to the string. We expect the restoring force to be proportional to the displacement for small oscillation. From above  = -Tsin= -  Where L is the length of the string and sin = x/L. The y-component of the acceleration is negligibly small, so  = Tcos – mg =may0 Since cos   1 for small, T  mg. Then   -  = max Solving for ax: ax = - x To identify the angular frequency, we recall that ax = -2x. Therefore, the angular frequency is  And the period is T =  = 2  There is a difference between angular frequency of the pendulum and it’s the velocity. Even though the two have the same units (rad/s in SI) and are written with the same symbol ω, for a pendulum they are not the same. When dealing with the pendulum, we use the symbol ω to stand for the angular frequency only. The angular frequency ω = 2 f of a given pendulum is constant, while the angular velocity changes with time between zero and its maximum magnitude. The results from the experiment is T= 1.34s and f= 0.75Hz b) Damping factors:- In harmonic motion, we assume that no dissipative forces such as friction or viscous drag exist, since the mechanical energy is constant, the oscillations contribute forever with constant amplitude. The oscillations of a swinging pendulum or a vibrating tuning fork gradually die out as energy is dissipated. The amplitude of each cycle is a little smaller than that of the previous cycle. This kind of motion is called Damped Oscillations, where the word damped is used in the sense of extinguished or restrained. For a small amount o damping, oscillations occur at approximately the same frequency as if there were no damping. A greater degree of damping lowers the frequency slightly. Even more damping prevents oscillations from occurring at all (Das and Saha, 1954). If the pendulum has mass m and the distance from the axis to the center f mass is d, then the torque is  = Fr = - (mg sin θ)d The component of the gravitational force, mg cos θ, passes through the axis of ration, so it does not contribute to the torque. In the equation above, both  and θ are positive if they are counterclockwise; the negative sign says that they always have opposite sign, since the torque always acts to bring θ close to zero. Assuming small amplitudes, sin θ ≈ θ (in radius) and the torque is  = mgd θ Thus, the restoring torque is proportional t to the displacement angle θ, just as the restoring force was proportional to the displacement for the simple pendulum and the mass on a spring. The net torque is equal to the rotational inertia times the angular acceleration:  = -mgdθ = 1 And the angular acceleration is  = -   Since the angular acceleration is a negative constant times the angular displacement from equilibrium, we indeed have SHM. ax = -ω2x Where  = ω2 Therefore, the angular frequency of the physical pendulum is ω2 = 2f =  and the period is T =  Where d is the distance form the axis to the CM and I is the rotational inertia. For a uniform bar of length L, the center of mass is halfway down the bar: d = ½ L the rotational inertia of a uniform bar rotating about an axis through an endpoint is I = ⅓ mL2. The period of oscillation is T = 2  Substituting for I and d, T=2  = 2 c) Calibration curves From the figure the gravitational force acts at the center of mass, but we cannot think of all the mass as being concentrated at that point-that would give the wrong rotational inertia. When set into oscillation, the bar, or any other rigid object free to rotate about a fixed axis. If the pendulum has torques about the rotation axis, only gravity gives a nonzero tongue. Resonance In amps Always circuit is connected to an ac source with a fixed amplitude but variable frequency. The impedance depends on frequency, so the amplitude of the current depends on frequency. The shape of these graphs is determined by the frequency dependence of the inductive and capacitive reactance. At low frequencies, the reactance of the capacitor Xc =1/(ωC) is much greater than either R or XL, so Z ≈ Xc. At high frequencies, the reactance of the inductor XL = ωL is much greater than either R or Xc, so Z ≈ XL. At extreme frequencies, either high or low, the impedance is larger and the amplitude of thee current is therefore small. The impedance of the circuit is Z =  Since R is constant, the minimum impedance Z = R occurs at an angular frequency ω0 – called the resonant angular frequency – for which the reactance of the inductor and capacitor are equal so that XL – Xc = 0. XL = XC ω0L =  Solving for ω0, ω0 =  Note that the resonant frequency of a circuit depends only on the values of the inductance and the capacitance, not on the resistances. In figure, the maximum current occurs at the resonant frequency for any value of R. However, the value of the maximum current depends on R since Z = R at resonance. The resonance peak is higher for a smaller resistance. If we measure the width of a resonance peak where the amplitude of the current has half its maximum value, we see that the resonance peaks get narrower with decreasing resistance. Resonance in a RLC circuit is analogous to resonance in mechanical oscillations. Just as a mass-spring system has a single resonant frequency, determined by the spring constant and the mass, the RLC circuit as a single resonant frequency, determined by the capacitance and the inductance. When either system is driven externally – by a sinusoidal applied force for the mass-spring or by a sinusoidal applied emf for the circuit – the amplitude of the system’s response is greatest when driven at the resonant frequency. In both systems, energy is being converted back and forth between two forms. For the mass-spring, the two forms are kinetic and elastic potential energy; for the RLC circuit, the two forms are electric energy stored in the capacitor and magnetic energy stores in the inductor. The resistor in the RLC circuit fills the role of friction in a mass-spring system: dissipating energy (Giambattista, Richardson and Richardson, 2007). Electrical resistance DISCUSSION Note that the period depends on L and g but not on the mass of the pendulum. Be careful not to confide the angular frequency of the pendulum with the angular velocity. Between though the two have the same units and the written with the same symbol for a pendulum they are the same. When dealing with the pendulum, we use the symbol to stand for the angular frequency only. The angular frequency of a given pendulum is constant, while the angular velocity, the rate of change of oscillations changes with time between zero and, maximum magnitude. When damping forces are present, the only way to keep the amplitude of oscillations from diminishing is to replace the dissipated energy from some other source. When a child is being pushed on a swing, the parent replaces the energy dissipated with a small push. In order to keep the amplitude of the motion constant, the parent gives a little push once per cycle, adding just enough energy each time to compensate for the energy dissipated in one cycle. The frequency of the driving force matches the natural frequency of the system. Driven oscillations occur when a periodic external driving force acts on a system that can oscillate. The frequency of the driving force does not have to match the natural frequency of the system. Ultimately, the system oscillates at the driving frequency, even it is far from the natural frequency. However, the amplitude of the oscillations is generally quite small unless the driving frequency f is close to the natural frequency of the system; the amplitude of the motion is a maximum. At resonance, the driving force is always doing positive work, the energy of the oscillator builds up until the energy dissipated balances the energy added by the driving force. For an oscillator with little damping, this requires large amplitude. When the driving and natural frequencies differ, the driving force and velocity are no longer synchronized. Sometimes they are in the same direction and sometimes in opposite directions. The driving force is not at resonance, so it sometimes does negative work. Therefore the oscillator’s energy and amplitude are smaller than at resonance(Eiichi Fukushima and Roeder, 1981). Large amplitude vibrations due to resonance can be dangerous in some situations. Materials can be stressed past their elastic limits, causing permanent deformation or breaking. Conclusion The numerical performances of these selected models have been evaluated in a virtual experiment relevant to the impact on shock absorbers and not completely negligible differences have been obtained considering the plastic strain levels in the target. These results indicate that the reliability of resonance impact numerical analyses on a structure could be improved performing different analyses and varying the modeling parameters within the ranges identified in the numerical–experimental correlation. Moreover, a correlation has been found between the time-averaged pressure at the centre of the impact obtained by oscillation on a rigid target, considering the total impact duration and the strain induced by force. Further numerical and experimental activities may investigate the application of the developed numerical approach in more complex impact conditions and the overall validity of the identified range of modeling parameters, also adopting different numerical approaches (De Jong, Kentgens, and Veeman, 1984). Reference List Das, T. P. and Saha, A. K. (1954) "Mathematical Analysis of the Hahn Spin-Echo Experiment," Phys. Rev. 93, 749-756. De Jong, A. F., Kentgens, A. P. M., and Veeman, W. S. (1984) "Two-Dimensional Exchange NMR in Rotating Solids: A Technique to Study Very Slow Molecular Reorientations," Chem. Phys. Lett. 109, 337-342. Eiichi Fukushima and Stephen B. W. Roeder, (1981) "Experimental Pulse NMR," Addison- Wesley, Reading, MA, 1981. Students love the approach. Giambattista, A., Richardson B.M. and Richardson, R.C.( 2007). College Physics. McGraw Hill Haeberlen, U. and Waugh, J. S. (1968) "Coherent Averaging Effects in Magnetic Resonance," Phys. Rev. 175, 453-467. Redfield, A. G. (1957) "On the Theory of Relaxation Processes," IBM Journal of Res. and Dev. 1, 19-31. Samoson, A., Lippmaa, E., and Pines, A. (1988) "High resolution solid-state N. M. R. Averaging of second-order effects by means of a double-rotor," Mol. Phys. 65, 1013-1018. Slichter, C. P. (1989). "Principles of Magnetic Resonance," 3rd Ed., Springer-Verlag, Berlin, Traficante, D. D. (1989) "Impedance: What It Is, and Why It Must Be Matched," Concepts in Magn. Reson. 1, 73-92. Waugh, J. S. and Huber, L. M. (1967) "Method for Observing Chemical Shifts in Solids," J. Chem. Phys. 47, 1862-1863. Read More

This expressed as shown below; Since the weight of the bob has no x-component, the restoring force is the x-component of the force due to the string. We expect the restoring force to be proportional to the displacement for small oscillation. From above  = -Tsin= -  Where L is the length of the string and sin = x/L. The y-component of the acceleration is negligibly small, so  = Tcos – mg =may0 Since cos   1 for small, T  mg. Then   -  = max Solving for ax: ax = - x To identify the angular frequency, we recall that ax = -2x.

Therefore, the angular frequency is  And the period is T =  = 2  There is a difference between angular frequency of the pendulum and it’s the velocity. Even though the two have the same units (rad/s in SI) and are written with the same symbol ω, for a pendulum they are not the same. When dealing with the pendulum, we use the symbol ω to stand for the angular frequency only. The angular frequency ω = 2 f of a given pendulum is constant, while the angular velocity changes with time between zero and its maximum magnitude.

The results from the experiment is T= 1.34s and f= 0.75Hz b) Damping factors:- In harmonic motion, we assume that no dissipative forces such as friction or viscous drag exist, since the mechanical energy is constant, the oscillations contribute forever with constant amplitude. The oscillations of a swinging pendulum or a vibrating tuning fork gradually die out as energy is dissipated. The amplitude of each cycle is a little smaller than that of the previous cycle. This kind of motion is called Damped Oscillations, where the word damped is used in the sense of extinguished or restrained.

For a small amount o damping, oscillations occur at approximately the same frequency as if there were no damping. A greater degree of damping lowers the frequency slightly. Even more damping prevents oscillations from occurring at all (Das and Saha, 1954). If the pendulum has mass m and the distance from the axis to the center f mass is d, then the torque is  = Fr = - (mg sin θ)d The component of the gravitational force, mg cos θ, passes through the axis of ration, so it does not contribute to the torque.

In the equation above, both  and θ are positive if they are counterclockwise; the negative sign says that they always have opposite sign, since the torque always acts to bring θ close to zero. Assuming small amplitudes, sin θ ≈ θ (in radius) and the torque is  = mgd θ Thus, the restoring torque is proportional t to the displacement angle θ, just as the restoring force was proportional to the displacement for the simple pendulum and the mass on a spring. The net torque is equal to the rotational inertia times the angular acceleration:  = -mgdθ = 1 And the angular acceleration is  = -   Since the angular acceleration is a negative constant times the angular displacement from equilibrium, we indeed have SHM.

ax = -ω2x Where  = ω2 Therefore, the angular frequency of the physical pendulum is ω2 = 2f =  and the period is T =  Where d is the distance form the axis to the CM and I is the rotational inertia. For a uniform bar of length L, the center of mass is halfway down the bar: d = ½ L the rotational inertia of a uniform bar rotating about an axis through an endpoint is I = ⅓ mL2. The period of oscillation is T = 2  Substituting for I and d, T=2  = 2 c) Calibration curves From the figure the gravitational force acts at the center of mass, but we cannot think of all the mass as being concentrated at that point-that would give the wrong rotational inertia.

When set into oscillation, the bar, or any other rigid object free to rotate about a fixed axis. If the pendulum has torques about the rotation axis, only gravity gives a nonzero tongue. Resonance In amps Always circuit is connected to an ac source with a fixed amplitude but variable frequency.

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